Problem 179 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}, 15 \mathrm{~mL}\) a gaseous hydrocarbon requires \(375 \mathrm{~mL}\) air containing \(20 \% \mathrm{O}_{2}\) by volume for complete combustion. After comustion the gases occupy \(330 \mathrm{~mL}\). Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: (a) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6}\)

Step-by-Step Solution

Answer

The formula of the hydrocarbon is \(\mathrm{C}_{4} \mathrm{H}_{10}\).

Step 1: Analyzing the Combustion Reaction

For a general hydrocarbon with formula \(C_xH_y\), the combustion reaction is \(C_xH_y + O_2 \rightarrow CO_2 + H_2O\). We are given 15 mL of the hydrocarbon and 375 mL of air (which is 20% \(O_2\)), so the volume of \(O_2\) is \(375 \times 0.20 = 75\) mL.

Step 2: Understanding Gas Volume Relation

Based on Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, the volume ratios directly reflect mole ratios.

Step 3: Write Balanced Combustion Equation

The balanced equation for a hydrocarbon combustion is \(C_xH_y + a O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). Based on 75 mL \(O_2\) being used for combustion of 15 mL hydrocarbon, we explore the stoichiometry for different hydrocarbons.

Step 4: Calculate Volume of Products

The total volume post-combustion is 330 mL, with the volume of gases being just \(CO_2\), because \(H_2O\) is in liquid form and does not contribute to the gas volume.

Step 5: Solve for Specific Hydrocarbon Formula

Let’s check each option. For \(C_3H_8\), 15 mL will consume 75 mL of \(O_2\) and produce 45 mL of \(CO_2\). However, for \(C_4H_{10}\), from the equation \(2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O\), 15 mL of \(C_4H_{10}\) needs 97.5 mL \(O_2\) and produces 60 mL \(CO_2\), aligning with post-combustion volumes. Hence, \(C_4H_{10}\) satisfies the condition.

Key Concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen, releasing energy in the form of heat and light. In the context of hydrocarbons, which are organic compounds composed entirely of hydrogen and carbon, combustion results in the formation of carbon dioxide (0_21) and water (H_2O2). These reactions are characterized by their exothermic nature, meaning they release heat. The general equation for the combustion of a hydrocarbon can be written as C_xH_y + O_2 ightarrow CO_2 + H_2O.

• The reactants involve a hydrocarbon and oxygen.
• The products are carbon dioxide and water.
• Typically, these reactions require a catalyst, such as heat or a spark, to get started.

Understanding combustion reactions helps in realizing how energy from fuels is harnessed and utilized in various engines and machinery. Knowing the stoichiometry, the quantitative relationships between reactants and products in a chemical reaction, assists in calculating the needed or produced volumes/masses of substances involved during combustion.

Avogadro's Law

Avogadro's Law is a fundamental principle in chemistry that relates the volume of a gas to the number of molecules or moles of the gas at a constant temperature and pressure. The law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. This simply means that if you have one liter each of oxygen and nitrogen gas at the same conditions, they will contain the same number of molecules, regardless of the gas type. Applied in stoichiometry, Avogadro's Law allows us to use volume ratios to deduce mole ratios in chemical reactions involving gases. For example, if you know the volume of oxygen used in a reaction, you can infer the number of moles of oxygen used. This is particularly helpful in combustion reactions, as it allows us to balance equations by comparing the volume of gases involved. In the context of our exercise, knowing the volumes of gases before and after a reaction, allows us to deduce the stoichiometry of the reactants and products, aiding in determining the empirical formula of unknown hydrocarbons.

Hydrocarbons

Hydrocarbons are compounds that consist solely of hydrogen and carbon atoms. They are categorized into several types depending on the types of bonds between the carbon atoms:

• Alkanes: These are saturated hydrocarbons with single bonds, as seen in methane (CH_4), ethane, and butane.
• Alkenes: Unsaturated hydrocarbons with at least one double bond between carbon atoms, like ethylene (C_2H_4).
• Alkynes: Also unsaturated, they have at least one triple bond between carbons, such as acetylene (C_2H_2).

Hydrocarbons are the primary constituents of fossil fuels, which means they are incredibly important for energy production. Their combustion releases energy, making them essential for various applications, from heating to powering engines. Understanding their chemical properties and reactions, like combustion, is crucial for fields like chemical engineering and environmental science.

Chemical Equation Balancing

Balancing chemical equations is a vital skill in chemistry that ensures the conservation of mass in a chemical reaction. According to the law of conservation of mass, the mass of reactants must equal the mass of products; therefore, we need identical numbers of each type of atom on both sides of the reaction equation. When balancing equations, each compound's formula must be fixed while you adjust the coefficients before each compound to achieve the balance.

• Identify the most complex molecule and start balancing with it.
• Balance the elements that appear in only one reactant and one product first.
• Save elements found in multiple compounds for last, like oxygen in combustion reactions.
• Check each element to confirm balanced status, adjusting coefficients where needed.

In the combustion of hydrocarbons, balancing typically starts with either carbon or hydrogen, concluding with oxygen. For example, the equation C_4H_{10} + O_2 ightarrow CO_2 + H_2O must be balanced by ensuring that the number of carbon, hydrogen, and oxygen atoms are equal on both sides of the equation. This ensures the reaction accurately represents matter conserved through the process, crucial for calculating precise reactant and product quantities.